The coordinate of a particle moving in a plane are given by ` x(t) = a cos (pt) and y(t) = b sin (pt)` where `a,b (lt a)` and `P` are positive constants of appropriate dimensions . Then

Options

A. The path of the particle is an ellipse.
B. The velocity and acceleration of the particle are normal to each other at ` t= pi//2p`.
C. The acceleration of the particle is always directed towards a focus.
D. The distance travelled by the particles in time interval ` t= 0 to t = pi//2p` is a.

Answer

A, B

Explanation

a,b. `x = a cos (pt), y = b sin (pt)`
Equation of path in x-y plane is
`[x/a]^2 + [y/b]^2 = 1`

i.e., the path of the particle is an ellipse.
Position vector of a point P is
`vec r = a cos pt hati + b sin pt hatj `
`rArr vecv = p(-a sin pthati + b cos pt hatj)`
and `veca = -p^2 (-a cos pt hati + b sin pt hatj) = -p^2vecr`
Acceleration of directed towards the centre.
Also , `vecv*veca = 0 at t = pi//2p`
At `t = 0 , vecr = a hati`
At ` t= pi//2p, vecr = bhatj`
So in time `t = 0 to t = pi//2p`, particle goes from A to B travelling
a distance more than a.